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Chapter 13: Problem 38

Differentiate implicitly to find the first partial derivatives of \(z\). $$ x \ln y+y^{2} z+z^{2}=8 $$

Short Answer

Expert verified
The first partial derivatives of \(z\) are \(\frac{\partial z}{\partial x} = - \frac{1}{4z}\) and \(\frac{\partial z}{\partial y} = - \frac{2xz}{y^2}\).

Step by step solution

01

Differentiating with respect to \(x\)

Differentiate implicitly each term in the equation with respect to \(x\). Apply the rule of logarithmic differentiation for term \(x \ln y\), product rule for \(y^{2} z\) and chain rule for \(z^{2}\).\n This yields \(\frac{1}{y} + 2yz \frac{\partial z}{\partial x} + 2z \frac{\partial z}{\partial x} = 0\). Simplifying gives \(\frac{\partial z}{\partial x} = - \frac{1}{4z}\).
02

Differentiating with respect to \(y\)

For differentiating implicitly with respect to \(y\), one should use the product rule for \(x \ln y\) as \(\frac{\partial(z)}{\partial(y)} = \frac{z}{y}\), for the term \(y^{2} z\) as \(2yz + y^{2} \frac{\partial z}{\partial y}\) and the chain rule for \(z^{2}\). This results in the equation \(x \cdot \frac{1}{y} + 2yz + y^2 \frac{\partial z}{\partial y} = 0\). From this, we get \(\frac{\partial z}{\partial y} = - \frac{2xz}{y^2}\).
03

Final solution

The first partial derivatives of \(z\), obtained by implicit differentiation, are \(\frac{\partial z}{\partial x} = - \frac{1}{4z}\) and \(\frac{\partial z}{\partial y} = - \frac{2xz}{y^2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Derivatives
Partial derivatives play an essential role in multivariable calculus, particularly when dealing with functions of more than one variable, such as the function in our exercise. Here's what you need to know about partial derivatives:
  • A partial derivative represents how a function changes as one of its variables changes, while the other variables are held constant.
  • When calculating a partial derivative with respect to a variable, we differentiate with respect to that variable and treat all other variables as constants.
  • In the context of the given exercise, we find the partial derivative of the function with respect to both variables, x and y, to understand how z changes in relation to each one.

Understanding these basics, we're better equipped to handle complex scenarios where functions don't just depend on a single variable, but on multiple interacting factors.
Chain Rule
The chain rule is a formula to compute the derivative of a composite function. When functions are nested with one inside another, the chain rule allows us to differentiate the entire function in a step-by-step manner. Here鈥檚 a brief overview:
  • The chain rule is written as \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \) where u is an intermediate function of x.
  • In the exercise, the term \(z^2\) required the application of the chain rule since z is a function of both x and y.
  • Understanding the chain rule is crucial for correctly differentiating complex functions where one variable is a function of another.
Product Rule
When differentiating products of functions, the product rule is the tool we use. It鈥檚 a rule that allows us to find the derivative of a product without having to expand the expression. Here's the gist of it:
  • The product rule is expressed as \(\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v鈥�(x)\).
  • In our exercise, the term \(y^2z\) was differentiated using the product rule because it is the product of two functions of y and z.
  • Learning the product rule is essential for dealing with multiplication within derivatives and avoids unnecessary complications of algebraic manipulation.
Logarithmic Differentiation
Logarithmic differentiation is a technique that simplifies the differentiation of complex functions involving exponents or products by taking the natural logarithm of both sides. This technique is particularly useful when the variables are multiplied by their own derivatives. Here's a breakdown:
  • Using the properties of logarithms, we can transform the product of functions into a sum, making differentiation easier.
  • For instance, in the term \(x \ln y\) from our exercise, logarithmic differentiation helps us to separate the variables x and y.
  • This method is powerful when differentiating functions where the direct application of fundamental rules would be cumbersome or overly complex.
The key to mastering logarithmic differentiation is understanding the properties of logarithms and their interaction with differentiation.

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Most popular questions from this chapter

Show that the tangent plane to the quadric surface at the point \(\left(x_{0}, y_{0}, z_{0}\right)\) can be written in the given form.Show that any tangent plane to the cone \(z^{2}=a^{2} x^{2}+b^{2} y^{2}\) passes through the origin.

Use Lagrange multipliers to find any extrema of the function subject to the constraint \(x^{2}+y^{2} \leq 1\). $$ f(x, y)=e^{-x y / 4} $$

Use Lagrange multipliers to find the minimum distance from the curve or surface to the indicated point. [Hint: In Exercise 23, minimize \(f(x, y)=x^{2}+y^{2}\) subject to the constraint \(2 x+3 y=-1 .]\) $$ \begin{array}{ll} \underline{\text { Surface }} & \underline{\text {Point}} \\ \text { Plane: } x+y+z=1& \quad(2,1,1) \end{array} $$

Find symmetric equations of the tangent line to the curve of intersection of the surfaces at the given point, and (b) find the cosine of the angle between the gradient vectors at this point. State whether or not the surfaces are orthogonal at the point of intersection.\(z=x^{2}+y^{2}, \quad z=4-y, \quad(2,-1,5)\)

A meteorologist measures the atmospheric pressure \(P\) (in kilograms per square meter) at altitude \(h\) (in kilometers). The data are shown below. $$ \begin{array}{|c|c|c|c|c|c|} \hline \text { Altitude, } h & 0 & 5 & 10 & 15 & 20 \\ \hline \text { Pressure, } P & 10,332 & 5583 & 2376 & 1240 & 517 \\ \hline \end{array} $$ (a) Use the regression capabilities of a graphing utility to find a least squares regression line for the points \((h, \ln P)\). (b) The result in part (a) is an equation of the form \(\ln P=\) \(a h+b\). Write this logarithmic form in exponential form. (c) Use a graphing utility to plot the original data and graph the exponential model in part (b). (d) If your graphing utility can fit logarithmic models to data, use it to verify the result in part (b).
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